The weights were not the argument

The Hex machine Shannon and Moore built in the early 1950s read the board as a network of resistors and moved where the potential pointed. A current through the board scored it against perfect play: it took a winning cell on more than 98 positions in a hundred and still lost every board it had to defend past two rows, and a circuit asked once found that making it commit to a table of replies did not help. The question those essays left was what the machine was missing. A pairing is checked, not measured brought the question down to a finite size. A winning strategy on boards of $n$ rows and $n + 1$ columns is a pairing: a set of disjoint pairs of cells such that every minimal chain Across could build contains a whole pair, so that Down, answering each of Across’s stones with its partner, is never crossed. No score that counts the chains a pair of cells shares has a winning pairing as its best on four by five. But a count allowed to listen did: after each best pairing, multiply the weight of every chain it missed by 1.5, and repeat. At round 46 the pairing it chose held the board, and it was the diagonal pairing, the one that has a sentence. Reflect each cell across the board’s diagonal and move one column.

That essay ended on the most hopeful reading of the result. If the loop converges on the reflection, then its final weights might be a numerical shadow of the argument, and reading them might say which chains the argument is really about. It is a historically interesting possibility, because it would mean that a measuring machine, given a way to learn from its failures, could find its way to the sentence — the kind of description where the needle has a sentence found for Chomp’s opening on exactly the families where the opening is a pairing. This essay tests it, and the answer is no. The weights are a point the loop happened to land on, not a place it was drawn to. What they do turn out to say is something narrower and more definite: chain weights can rule a pairing out, and nothing short of being told the answer can rule one in.

One rate in twelve #

The first test of whether a result is an attractor is to nudge the thing that produced it. The loop has one parameter, the factor by which a missed chain’s weight grows. A pairing is checked, not measured tried three, 1.25, 1.5 and 2, and found that only the middle one held four by five. Three is a small sample, and two of three failing could mean the good rates are a broad band with 1.5 inside it and the others just outside.

They are not. Run at twelve rates from 1.1 to 3, a hundred rounds each, the loop holds four by five at ×1.5 and at no other rate. At 1.4 and at 1.6, on either side, it stalls with four chains and with two still missed. At 1.1, the gentlest, the best it manages is two missed chains; at 2, 2.5 and 3 it swings and never settles below four. There is no band, only the one value that was tried first in the earlier essay because it sat between the others.

The small boards say nothing either way, and the picture shows why. Two by three has five minimal chains and the chain count’s first pairing already holds it. Three by four has 25, the first pairing misses one, and raising that one chain by any factor at all moves the best pairing to the diagonal at the second round. These boards cannot distinguish an attractor from luck, because there is nothing to steer: one correction does the whole job, and it does it at every rate. Four by five, with 148 chains, is the first board where the loop has to find its way, and there it finds its way exactly once.

One start in twenty #

The second nudge is to the starting point. The loop began with every chain at weight one, which is the natural start and also an arbitrary one. A loop that is drawn toward the diagonal should find it from nearby starts.

Give each of the 148 chains a first weight drawn at random between a half and one and a half, seeded so the draw can be repeated, and run the same loop at the same rate. One start in twenty holds the board, at round 63, and it holds it with the diagonal pairing again. The other nineteen stall. One gets as close as a single missed chain, the rest sit at two, three or four.

The single success is worth reading carefully, because it pulls in the opposite direction from the failures. When the loop holds four by five at all, from any start or rate tried, it holds it with the diagonal, never with either of the two other winning pairings that three pairings and no order found. So there is something about the diagonal that the loop prefers. But that preference is visible only in the rare runs that succeed, and twenty starts produce one of them. Whatever pulls the loop toward the diagonal is weak beside whatever stalls it, and the run from equal weights was a trajectory that happened to arrive, not the bottom of a basin that every nearby trajectory falls into.

What the loop had to raise #

The run that did arrive can still be read, and the earlier essay’s question was specifically about its weights: which chains it raised, by how much, and whether they are the chains crossing the diagonal where the reflection argument uses them.

In 46 rounds the loop raised 71 of the 148 chains, the most-raised ten times, so that chain ends at 1.5 to the tenth, about 58 times its starting weight. The reflection argument has a natural division of the chains to compare against. Every chain contains at least one whole diagonal pair, which is what it means for the diagonal pairing to hold. Seventy-one chains contain exactly one: on those the argument has no slack, and Down’s whole defence rests on that single pair. The rest contain two, three or, on one chain, four, and the argument could lose a pair on any of them and still hold.

If the weights were the argument in numbers, they would sit on the 71 tight chains and leave the others alone. They lean that way. Forty-five of the 71 tight chains were raised, against 23 of the 66 held by two diagonal pairs and two of the ten held by three. But 26 of the raised chains are ones the diagonal holds with a pair to spare, and one of them was raised ten times, as often as any chain on the board. The length grouping says more plainly what the loop was responding to. It raised 26 of the 32 shortest chains, those of five cells, and none of the 18 longest, those of nine. Short chains are the ones with the fewest cells, so a pairing that pairs a few of their cells elsewhere leaves them empty most easily, and those are the misses the loop sees. It was correcting for which chains its own pairings tended to miss, which is a fact about the chain count’s habits, not about the diagonal.

The half-turn gives one further check. Turning the board through 180 degrees carries Across’s chains to Across’s chains and carries every one of the three winning pairings to itself, so a set of weights that encoded any of them faithfully should be unchanged by it. The loop’s weights nearly are: 146 of the 148 chains were raised exactly as often as their images, and two differ by one raising, which is the trace of the pairing search breaking a tie in one direction rather than the other. The symmetry is respected. It also tells nothing about the diagonal in particular, since the other two winners respect it equally.

A narrow place to stand #

A point that the loop arrived at by luck could still be a broad region once it is there. The last robustness test perturbs the destination instead of the start.

Multiply each final weight by a random factor within a stated band and take the heaviest pairing again. Within one per cent and within five, the diagonal survives every draw. Within ten it survives nineteen of twenty, and within twenty per cent eleven. That is a modest region, wide enough that the result is not an accident of floating-point arithmetic and narrow enough that a fifth’s uncertainty in the weights loses the board about half the time. What the noise shows besides is the shape of the alternatives. Whenever the diagonal is not the heaviest pairing, the heaviest pairing loses. The weights have not found a region in which winning pairings are preferred and the diagonal happens to be the best of them. They sit just inside the edge of the diagonal’s own region, with losing pairings on the other side of it.

Knowing which chains matter is not enough #

The robustness tests ask whether the loop’s weights are stable. The sharper form of the original question asks whether any weights that do encode the argument would work, whatever the loop happened to find. The tight chains are the obvious candidate: if the argument is about anything, it is about the chains it holds with a single pair. So take each winning pairing in turn, give weight one to every chain it holds with exactly one pair and nought to the rest, and find the heaviest pairing under those weights.

It never returns the pairing the chains were taken from, or any other winner. On three by four each winner’s tight chains give a heaviest pairing that misses one chain. On four by five the diagonal’s 71 tight chains give a pairing that misses four. The second winner’s 82 give one that misses 22, and the third’s 115 give one that misses six. Weighting the chains the argument relies on is not a measurement that recovers the argument, and the reason is the arithmetic a pairing is checked, not measured found under every chain score. A heaviest pairing maximises a sum, and a sum is content to put two pairs on one heavy chain and none on another. A pairing holds by putting at least one on every chain, which is a minimum, and weighting the right chains does not turn a sum into a minimum.

That closes the hopeful reading. The loop’s final weights are not the tight chains, they are not robust, and even the tight chains as weights would not have worked. The weights that held four by five are a particular vector that the loop’s particular trajectory reached, and there is no sentence in them to read.

Taught, every winner is reachable #

That leaves a question the negative raises about itself. Perhaps the diagonal is the only winning pairing that any chain weighting can make heaviest, and the loop finds it on the rare occasions it finds anything because it is the only thing there to find. That would be a real fact about the diagonal, and it can be tested directly: hand a measurement the answer and see whether it can be made to agree.

The procedure is a perceptron over chain weights. Hold a target pairing, find the heaviest pairing that is not the target, and if that rival weighs as much as the target or more, move every chain’s weight up by the number of whole pairs the target has on it and down by the number the rival has, never below nought. Repeat until the target is strictly heaviest. On four by five all three winning pairings can be made heaviest: the diagonal after six corrections, the second winner after four and the third after twelve. The diagonal is not privileged. Every winner sits in some region of weight space where it is the best pairing, and a measurement that knows where it is going gets there in a dozen steps, where the blind loop needed 46 rounds at its one working rate and usually never arrived.

That is the difference the historical machines were missing, and it has a name older than this question. The Nim machine of 1940 never faced it, because it was built around a theorem and had nothing to learn. Rosenblatt’s perceptron, built at Cornell in the late 1950s, adjusted weights after each mistake toward the right answer, and the theorem that it converges, proved in 1962, needs exactly that: a right answer to move toward, and the guarantee that some weighting separates it from everything else. The listening loop is told only that it was wrong and where, a chain left empty, and not what would have been right. Being told where it failed is enough on three by four, where one failure points straight at the fix. On four by five it is a search with a weak sense of direction, and the perceptron told the target is a different procedure: it is checking an answer it already has. So the machine was not missing a better measurement. A measurement that could hold the board would have to be taught the answer, and the answer is the certificate.

A pairing measurement cannot reach #

The perceptron does have one failure, and it is the most definite result here.

On three by four the pairing that joins the two far corners of the board, top left with bottom right, holds the board as surely as the other two. Told to prefer it, the perceptron stops at once, because the best rival is another winning pairing that has at least as many whole pairs on every one of the 25 chains. Both other winners have that property. A measurement of the kind every chain score and every circuit reading has been scores a pairing by adding up, chain by chain, a weight for each whole pair. If another pairing matches or beats the corner pairing on every chain, it matches or beats it in every such sum, whatever the weights are. So the corner pairing is a winning strategy that no chain measurement can ever choose. Comparing every one of the 10,395 pairings of three by four against each winner, and every one of the 15 on two by three, it is the only shape that happens: on two by three the corner-to-corner winner is covered the same way by the diagonal, and nothing rules out any other winner.

That is a rule-out, and it is the strongest thing chain weights can establish. Dominance is checkable and final: a pairing covered chain by chain by another is out of reach of every weighting at once, and this is a fact about the board’s chains, not about any trajectory. On four by five nothing is ruled out this way. None of the three winners covers another chain by chain, and the perceptron reaches all three. On the small boards the corner pairing is the casualty. It is the pairing a measurement can never see, and it wins anyway. The same split runs through a potential that names every move and the winning reply is the fourth choice, where a weighting of Across’s chains in play held every three-row board Down can win and lost four by five, on which the winning reply was on average its fourth choice: a sum over chains can be tuned to agree with a strategy and cannot be relied on to find one. The theorem that names no move is the far end of the same line, a proof that a winning strategy exists with no measurement in it at all.

The convention every count rests on #

Hex on rhombic boards of $n$ rows and $n + 1$ columns, Down joining the nearer pair of edges and moving second, as in the essays before this one. A minimal chain is a set of cells joining Across’s two edges from which no cell can be removed; a pairing holds when every minimal chain contains both cells of some pair. “Heaviest” means the pairing whose pairs’ weights add to the most, found exactly over all pairings by dynamic programming, with ties broken the same way every time. The loop starts its weights at one unless a start is stated, multiplies each missed chain’s weight by the rate, and stops at a hundred rounds. Seeded draws use a fixed linear congruential generator, so every count here can be repeated. “Covers every chain at least as often” compares the number of whole pairs a pairing has on each chain, one chain at a time.

What the weights cannot show #

That no rate or start outside those tried holds four by five more reliably. Twelve rates and twenty starts were run, and a narrow band of good rates between two tested values would not have been found. The finding is that the band, if it exists, is narrow, not that it is empty.

Anything past four by five. Five by six has thirty cells, past the pairing tables used here. Whether the loop’s rare successes still land on the diagonal there, and whether any winner there is ruled out the way the corner pairing is on three by four, is unmeasured.

Why the diagonal is the one the loop reaches when it reaches anything. Two successes on four by five, both diagonal, plus the small boards: it is a pattern, and the robustness tests say it is a weak one. The perceptron says it is not because the diagonal is the only winner a weighting can reach.

Still open: whether dominance is the whole of it #

The corner pairing on three by four is out of every measurement’s reach because another winner covers it chain by chain. That is one way to be unreachable, and it is not the only possible one. A pairing could also be out of reach because some mixture of other pairings, averaged, covers it chain by chain, while no single one does, and on four by five, where the perceptron reached all three winners, that never happened. The question that remains is whether the two coincide in general: whether, on every board of this shape, the winning pairings a chain measurement can never choose are exactly those another winner covers one chain at a time. If they do, then which strategies a measuring machine can ever find is decided by a comparison a reader can check pair by pair. The measurement is the same perceptron on five by six, with its rival search widened to pairings that average others, and it needs the chain list of a thirty-cell board, which is where the tables stop.